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Math Shortcuts
The radii of circumscribed (R) and inscribed (r) circles within regular polygons of n sides, each of length x, are given by:
(x/2) csc(180°/n)
and
(x/2) cot(180°/n)

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Previous Puzzles of the Month + Solutions

 
 
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Send to a Friend Puzzle # 125
puzzle no. 125

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story
 helm of MarsA mathematical shield
  
Once upon a time, Mars, the God of war, intended to test the IQ of the goddess Minerva. So, showing his shield, he told her: "Darling, on my shield, there are 3 equal circles which represent the qualities of the warrior: strength, flexibility, and decisiveness. As you can see, one of the circles has been scratched by a sword, and the resulting score is three inches long (line AB in the illustration). Can you then tell me what is the area of my shield?".
Find the very shortest way to solve this puzzle and use only basic geometry, trigonometry is not allowed!

Difficulty level: bulbbulb, basic geometry knowledge.
Category: Geometrical puzzle.
Keywords: inscribed circles, incircles.
Related puzzles:
- Red monad,
- Achtung Minen!


Source of the puzzle:
© G. Sarcone.
You cannot reproduce any part of this page without prior written permission.
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Solution

The centers of the small green circles are equidistant from each other, and thus form the vertices of an equilateral triangle with midpoints A, B and C (see image below).
It follows that the small triangles ABC and AC'B are also equilateral and identical to each other. The radii r of the small circles are congruent to the sides of the triangles ABC and AC'B, then r = C'B = AB.

As shown in the image, the radius R of the large circle can be calculated by adding r + h + NM together.

We can calculate the height h of the triangle AC'B by multiplying its side AB by √3/2:
h = (AB√3)/2 = (3√3)/2

Since the heights of an equilateral triangle meet at a point (here, point M) that is two thirds of the distance from the vertex of the triangle to the base, we can also calculate MN with this simple formula:
MN = h x 1/3 = (3√3)/2 x 1/3 = (3√3)/2 x 3 = √3/2

Therefore, the area of large yellow circle (which represents the shield) is:
π[3 + (3√3)/2 + √3)/2]2 = π(3 + 2√3)2131.27 square inches

puzzle solution


cup winnerThe 5 Winners of the Puzzle of the Month are:
John Pelot, USA USA flag - Benoît Humbert, France French flag - Amedeo Squeglia, Italy Italian flag - Paritosh Singh, India Indian flag - Walter Jacobs, Belgium Belgian flag

Congratulations!

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Math fact behind the puzzle
Properties of the equilateral triangle
An equilateral triangle is simply a specific case of a regular polygon, in this case with 3 sides.
Equilateral triangles are triangles in which all sides are equal, and all angles are equal as well and each of them measures 60.
With an equilateral triangle, the radius of the incircle is exactly half the radius of the circumcircle.

 

© 2006 G. Sarcone, www.archimedes-lab.org
You can re-use content from Archimedes’ Lab on the ONLY condition that you provide credit to the authors (© G. Sarcone and/or M.-J. Waeber) and a link back to our site. You CANNOT reproduce the content of this page for commercial purposes.

You're encouraged to expand and/or improve this article. Send your comments, feedback or suggestions to Gianni A. Sarcone. Thanks!
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Previous puzzles of the month...
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Solved Puzzles

Apr-May 2010:
A chopping problem
Oct-Nov 09:
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July-Sept 09:
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May-June 09:
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Jan-Feb 09
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Sept-Oct 08
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July-Aug 08:
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May-June 08:
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Febr-March 08
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Dec 07-Jan 08
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Oct-Nov 2007
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Aug-Sept 07:
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June-July 07:
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April-May 07:
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Febr-March 07:
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Jan 07:
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Aug-Sept 2006
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June-July 06
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Apr-May 06:
intriguing probabilities

Febr-March 06:
cows & chickens
Dec 05-Jan 06:
red monad

Sept-Oct 2005
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magic star

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